Math 527 - Homotopy Theory
Spring 2013
Homework 10 Solutions
Problem 1. Let n ≥ 2 [Sorry I forgot to write that on the homework!]. Consider the wedge
X = S 1 ∨ S n.
a. Show that the nth homotopy group of X is a free π1 (X)-module on one generator:
πn (X) ∼
= Z[π1 (X)] ∼
= Z[t, t−1 ].
Solution. By van Kampen, the fundamental group of X is the free product
π1 (X) ∼
= π1 (S 1 ) ∼
= Z.
= π1 (S 1 ) ∗ π1 (S n ) ∼
e of X consists of the real line R with a sphere Sin wedged at each integer
The universal cover X
e so that
i ∈ Z ⊂ R. The fundamental group π1 (X) ∼
= Z acts via deck transformations on X,
∼
=
n
via the identity function, and translates the real line
→ Si+k
k ∈ π1 (X) shifts the spheres Sin −
R by k, i.e. k · x = x + k for x ∈ R.
' W
e produces a homotopy equivalence X
e−
Collapsing the line R ⊂ X
→ i∈Z Sin where the induced
action of π1 (X) shifts the wedge summands. As an abelian group, we have
!
_
πn
Sn ∼
= Zhλi |i ∈ Zi
i
i∈Z
W
the free abelian group on all summand inclusions λi : Sin ,→ j∈Z Sjn . The π1 (X)-action is given
by
k · λi = λi+k
W
so that πn i∈Z Sin is a free π1 (X)-module on one generator, say λ0 . To conclude, note that
∼
=
e → X induces an isomorphism p∗ : πn (X)
e −
the covering map p : X
→ πn (X).
1
b. Take a representative f : S n → X of the class 2t − 1 ∈ πn (X) and form a space Y by
attaching an (n + 1)-cell to X via f , as illustrated in the cofiber sequence:
f
j
Sn →
− X→
− Y.
j
ι
1
Show that the composite S 1 −
→
X→
− Y induces an isomorphism on integral homology:
'
H∗ (S 1 ; Z) −
→ H∗ (Y ; Z).
Here ι1 : S 1 ,→ X is the wedge summand inclusion.
e → X becomes
e ' W Sin , the covering map p : X
Solution. Via the homotopy equivalence X
i∈Z
the fold map followed by the summand inclusion
W
∇
n
i∈Z Si
/
ιn
Sn
/
S 1 ∨ S n.
e → X induces the “fold” map on integral homology
Therefore p : X
p∗
/
e
Hn (X)
∼
=
Hn
Hn (X)
n
i∈Z Si
W
/
Hn (S 1 ∨ S n )
O
O
∼
=
L
∼
=
∇
n
i∈Z Hn (Si )
P
mi ui /
/
Hn (S n )
P
( mi )u
where ui ∈ Hn (Sin ) ' Z and u ∈ Hn (S n ) are suitably chosen generators, u being chosen for the
n-cell of Y .
∼
=
Here, the isomorphism Hn (S n ) −
→ Hn (S 1 ∨ S n ) is induced by the summand inclusion ιn : S n ,→
S 1 ∨ S n . Its inverse is induced by the summand collapse map cn : S 1 ∨ S n S n .
The (homotopy) commutative diagram
2ι1 −ι0
;
e
X
/
Sn
p
/
S1 ∨ Sn
cn
f
2
Sn
induces the following diagram on integral homology:
8
2u1 −u0
e
Hn (X)
/
n
Hn (S )
1
p∗
n
Hn (S ∨ S )
f∗
∼
=
cn∗
/
v /
Hn (S n )
(2 − 1)u
where again v ∈ Hn (S n ) is a generator, chosen for the (n + 1)-cell of Y .
With these choices of generators, the cellular chain complex of Y contains the differential
1
CW
Cn+1
(Y ) ' Z →
− Z ' CnCW (Y )
and thus the homology of Y is isomorphic to that of S 1 .
It remains to check that the map j ◦ ι1 : S 1 → Y induces an isomorphism on integral homology.
ι1
→
S 1 ∨ S n induces an
The only case to check is H1 . Note that the summand inclusion S 1 −
isomorphism on H1 :
H1 (S 1 ) → H1 (S 1 ∨ S n ) ∼
= H1 (S 1 ) ⊕ H1 (S n ) ∼
= H1 (S 1 ).
Moreover, j : X ,→ Y is the inclusion of the n-skeleton, and thus an n-connected map. Therefore
j induces an isomorphism on homology Hk in dimensions k < n, in particular in dimension
k = 1 < n.
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Alternate solution for that last step. Since j ◦ ι1 is the inclusion of a subcomplex, in
particular a cellular map, it induces a map of cellular chain complexes C∗CW (S 1 ) → C∗CW (Y )
which can be explicitly written as follows:
dimension
C∗CW (S 1 )
n+1
0
/
/
Z
'
n
C∗CW (Y )
/
0
Z
/
0
..
.
..
.
/
Z
1
0
1
/
Z
0
0
0
0
1
1
Z
/
0
Z
'
→ H∗CW (Y ; Z).
and clearly induces an isomorphism on homology H∗CW (S 1 ; Z) −
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c. Show that the same map j ◦ ι1 : S 1 → Y induces an isomorphism on πk for k < n but not
on πn .
Solution. Since j ◦ ι1 : S 1 → Y is the inclusion of the (n − 1)-skeleton of Y , it is an (n − 1)connected map. Moreover, we have:
πn−1 (Y ) ∼
= πn−1 (Yn )
= πn−1 (S 1 ∨ S n )
∼
e
= πn−1 (X)
=0
so that j ◦ ι1 : S 1 → Y induces (trivially) an isomorphism on πn−1 .
It remains to check πn (Y ) 6= 0.
Since X is path-connected and Y is obtained from X by attaching an (n + 1)-cell via the
attaching map f : S n → X, the resulting homotopy group πn (Y ) is the quotient of πn (X) '
Z[t, t−1 ] by the π1 (X)-submodule generated by f = 2t − 1 (c.f. Botvinnik Theorem 11.1).
The ring map
1
: Z[t, t−1 ] → Z[ ]
2
1
t 7→
2
is well defined and makes Z[ 21 ] into a Z[t, t−1 ]-module. Moreover, is clearly surjective, and one
readily checks the equality ker = (2t − 1) so that induces the isomorphism
1
Z[t, t−1 ]/(2t − 1) ' Z[ ].
2
In particular, we conclude πn (Y ) ' Z[t, t−1 ]/(2t − 1) 6= 0 as claimed.
5
Problem 2. (May § 15.2 Problems 3 and 4) Let n ≥ 1 and let G be an abelian group.
a. Construct a connected CW complex X whose reduced integral homology is given by:
(
e i (X; Z) ' G if i = n
H
0 if i 6= n.
Such a space is called a Moore space and is denoted M (G, n).
Solution. Let
ϕ
0 → F1 −
→ F0 → G → 0
be an exact sequence of abelian groups where Fi are free abelian groups. We used the fact that
a subgroup of a free abelian group is always free.
Realize ϕ as the homology of a map between wedges of spheres. More precisely, given isomorphisms
M
F1 ' Zhgi |i ∈ Ii ∼
Z
=
i∈I
F0 ' Zhhj |j ∈ Ji ∼
=
M
Z
j∈J
let us build a map
_
f:
Sin →
_
Sjn
j∈J
i∈I
satisfying Hn (f ) = ϕ (via the isomorphisms above). The restriction of f to the wedge summand
Sin is chosen so that its homotopy class satisfies
∈
f|
Sin
πn
_
W
n
j∈J Sj
Hurewicz
∈
ϕ(gi )
Hn
W
L
n
S
' j∈J Z
j∈J j
which is always possible since the Hurewicz morphism for wedges of spheres S n is surjective
(and in fact an isomorphism if n ≥ 2).
Call the wedges of spheres A and B respectively, and let X be the cofiber of f : A → B. Then
f
the long exact sequence on homology of the cofiber sequence A →
− B → X yields
/
...
/
e k (B)
H
/
e k (X)
H
/
e k−1 (A)
H
...
e k (X) = 0 for all k 6= n, n + 1. In the critical dimensions, the exact sequence is
which proves H
e n+1 (B)
H
/
e n+1 (X)
H
O
'
0
/
ker(ϕ)
/
f∗
e n (A)
H
O
/
e n (B)
H
O
'
/
/
O
F1
6
/
/
e n−1 (A)
H
'
'
ϕ
e n (X)
H
F0
/
coker(ϕ)
/
0
e n+1 (X) = 0 and H
e n (X) ' G. Therefore X = M (G, n) is a Moore space and
which proves H
has been built as a connected CW complex.
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b. Construct a connected CW complex Y whose homotopy groups are given by:
(
G if i = n
πi (Y ) '
0 if i 6= n.
Such a space is called an Eilenberg-MacLane space and is denoted K(G, n).
Solution.
Case n ≥ 2. The Moore space M (G, n) in part (a) was explicitly built as a complex with a
single 0-cell and cells in dimensions n and n+1, so that the its 1-skeleton is trivial and therefore
so is its fundamental group π1 M (G, n).
By the Hurewicz theorem, the bottom homotopy groups of M (G, n) are
πi M (G, n) = 0 if i < n
πn M (G, n) ∼
= Hn (G, n) = G.
Therefore the Postnikov truncation Y = Pn M (G, n) is an Eilenberg-MacLane space K(G, n).
Moreover, recall that the Postnikov truncation W → Pn W can be obtained by attaching cells
(of dimension at least n + 2). This produces a model for K(G, n) which is a connected CW
complex.
Case n = 1. The construction from part (a) must be adapted when n = 1. Let
ϕ
1 → F1 −
→ F0 → G → 1
be an exact sequence of groups where Fi are free groups. We used the fact that a subgroup of
a free group is always free.
Realize ϕ as π1 of a map between wedges of circles. Given isomorphisms
F1 ' F hgi |i ∈ Ii ∼
= ∗i∈I Z
F0 ' F hhj |j ∈ Ji ∼
= ∗j∈J Z
where F hSi denotes the free group on a set S of generators, let us build a map
_
_
Sj1
f:
Si1 →
i∈I
j∈J
satisfying π1 (f ) = ϕ (via the isomorphisms above). This is always possible, because of the
isomorphism
!
_
π1
Sj1 ∼
= F hιj |j ∈ Ji
j∈J
where ιj : Sj1 ,→
W
k∈J
Sk1 denotes the summand inclusion.
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Call the wedges of circles A and B respectively, and let X be the cofiber of f : A → B. Then X
is a CW complex with a single 0-cell. Hence, applying π1 to the cofiber sequence A → B → X
yields the exact sequence
π1 (A)
O
'
/
/
π1 (B)
O
/
1
/
1
O
'
F1
/
π1 (X)
'
/
F0
G
which proves π1 (X) ' G. The Postnikov truncation P1 X is an Eilenberg-MacLane space
K(G, 1), which moreover has been built as a connected CW complex.
Remark. Applying P1 to the Moore space M (G, 1) would not work in general, because π1 M (G, n)
is not abelian in general. For example, with G = Z ⊕ Z, a space M (Z ⊕ Z, 1) obtained in part
(a) could be S 1 ∨ S 1 . However, its fundamental group π1 (S 1 ∨ S 1 ) ∼
= F2 is the free group on
two generators, which is highly non abelian.
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Problem 3. Let n ≥ 0 and let X be a space with the homotopy type of a CW complex.
Consider the Postnikov truncation map tn : X → Pn X, which may be assumed a relative CW
complex.
Let f : X → Z be any map, where Z is an Eilenberg-MacLane space of type (G, k) for some
abelian group G and k ≤ n.
Show that there exists a map g : Pn X → Z satisfying f ' g ◦ tn , and this map g is unique up
to homotopy. Here, g makes the diagram
tn
X
/
Pn X
g
!
f
Z
commute up to homotopy.
Solution. We want to show that restriction along tn : X → Pn X induces a bijection on
homotopy classes of maps:
t∗n : [Pn X, Z] → [X, Z].
Since the functor [−, Z] is invariant under homotopy equivalences, WLOG X is a CW complex
(and tn : X → Pn X can still be assumed a relative CW complex).
Since X and Pn X are CW complexes, mapping into an Eilenberg-MacLane space K(G, k)
computes cohomology:
t∗n
/
[Pn X, Z]
θPn X
∼
=
∼
=
H k (Pn X; G)
[X, Z]
/
θX
H k (X; G).
t∗n
Here we used the fact that G is abelian to cover the case k = 1 as well:
[X, K(G, 1)] ∼
= H 1 (X; G)/conjugation in G
∼
= H 1 (X; G).
Note that tn : X → Pn X is an (n + 1)-connected map, and thus induces isomorphisms on
homology and cohomology with any coefficients in degrees less than n + 1. In particular, given
k ≤ n, the restriction map
'
t∗n : H k (Pn X; G) −
→ H k (X; G)
is an isomorphism. Therefore the restriction map
'
t∗n : [Pn X, Z] −
→ [X, Z]
is also a bijection (in fact an isomorphism of abelian groups).
Remark. The statement still holds when Z is a product of such Eilenberg-MacLane spaces.
However, if Z is more complicated, but still n-truncated (i.e. πi (Z) = 0 for i > n), then such a
factorization g : Pn X → Z still exists, but its homotopy class need not be unique.
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